Usually (eg, intro. in M. Rost's 'Cycle modules with coefficients'), for a variety, $X$, over a field one can define the Chow group of p-cycles, $CH_p (X)$, as
$$CH_p (X) = coker\; \left[\bigoplus_{x\in X_{p+1}} k(x)^\times \rightarrow \bigoplus_{x\in X_p} \mathbb{Z}\; \right]$$.

What about for an arithmetic scheme, eg when $X$ is, say, normal, separated, of finite type and flat over $Spec \; \mathbb{Z} $? Does something go wrong with the above definition?

Peter Arndt had posed part of this question already, but it seems without an answer.

5 Answers
5

One can define Chow groups over any Noetherian scheme $X$. Let $Z_iX$ be the free abelian group on the $i$-dimensional
subvarieties (closed integral subschemes) of $X$. For any $i+1$-dimensional
subscheme $W$ of $X$, and a rational function $f$ on $W$, we can
define an element of $Z_iX$ as follows:
$$ [div(f,W)] = \sum_{V} ord_V(f)[V] $$
summing over all codimension one subvarieties $V$ of $W$. Then the $i$-Chow group $CH_i(X)$ is defined
as the quotient of $Z_iX$ by the subgroup generated by all elements of the form $[div(f,W)]$. The order function is defined as length of the corresponding local ring, so it does not need any further assumptions on $X$, In fact, $X$ being locally Noetherian is enough. This definition also agrees with the one in the original question.

As explained by Hailong, one can define the Chow group CH($X$) (without grading by the dimension of the cycles) of any Noetherian scheme $X$. But in general one has to be careful about the behavior of the rational equivalence.

A principal divisor in an integral closed subscheme $W$ of dimension $i+1$ is not necessary a $i$-cycle (this has to do with the notion of catenary schemes).

If $X$ is not equidmensional, then the divisor associated to an invertible rational function can be not rationally equivalent to $0$ (even when $X$ is a reduced projective variety over a field).

If $f : X\to Y$ is a proper morphism, then contrarily to the case of varieties over a field, the pushforward by $f$ does not induce a map ${\rm CH}(X)\to {\rm CH}(Y)$. One can construct an example with $X$ affine, regular of dimension $2$, $f$ finite birational and the image of a principal divisor on $X$ is non-zero in CH($Y$).

To remedy to these pathologies, Thorup (Rational equivalence theory on arbitrary
Noetherian schemes, Enumerative geometry (Sitges, 1987), 256--297,
Lecture Notes in Math., 1436, 1990) defines a graded rational equivalence relation associated to a grading on $X$ (a map from $X$ to $\mathbb Z$ with some properties, e.g. $x\mapsto -\dim O_{X,x}$), and a graded principal divisor on an integral closed subscheme is the usual principal divisor where we discount components of bad gradings. With the new rational equivalence everything works fine for the pushforward by proper morphisms $X\to Y$ (though the gradings on $X$ and $Y$ must be compatible in some sense) and pullback by flat morphisms.

I learned most of these from O. Gabber and the examples mentioned above are in a (not yet finished) preprint with him and D. Lorenzini.

[EDIT]. To summarize, if $X$ is noetherian, universally catenary (e.g. finite type over ${\mathbb Z}$ or any noetherian regular scheme) and equidimensional at every point (i.e. for every $x\in X$, the irreducible components of ${\rm Spec}(O_{X,x})$ all have the same dimension), then CH$(X)$ can be decomposed as the direct sum of the $CH_i(X)$'s as in Hailong's post. If $f : X\to Y$ is a proper morphism of universally catenary noetherian schemes, then $f_*$ induces a homomorphism ${\rm CH}(X)\to {\rm CH}(Y)$. In the last counterexample above, $X$ is regular (so univ. catenary), $Y$ is catenary but not universally.

Thank you for the illuminating examples and references Prof. Liu, I am looking forward to the Gabber/Lorenzini article.
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IvanMar 9 '10 at 22:48

1

Hey I am one of the authors :). You might have a look at a paper of Shuji Saito and K. Sato to appear in the Annals : <i>A finiteness theorem for zero-cycles over $p$-adic fields</i> (also in Arxiv. They use CH(X) over ${\mathbb Z}_p$.
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Qing LiuMar 10 '10 at 12:55

So with a lot of extra care about dimension/codimension it seems to be possible to define Chow groups over Spec Z if I understand the above answers correctly.

I may point out that in the book by Elman, Karpenko and Merkurjev "Algebraic and Geometry Theory of Quadratic Forms" (even though the title does not suggest so) they very carefully work out Chow groups, even some version of higher Chow groups. They begin by treating Chow groups over general excellent schemes (something you do not have written so explicitly in Fulton), so quite general and only later impose additional assumptions, like equidimensionality, being over a field, and all that. So maybe it is worth having a look at that.

On the other hand, they get a pullback along non-flat morphisms only with the typical more restrictive conditions. This however is crucial for turning Chow groups into the Chow ring.

So I think the construction of the intersection product [which uses the pullback along the embedding of the diagonal X -> X x X] is another very very critical matter over Z (but according to one of the other answers it can be done, that sounds very interesting).

Last but not least, just maybe another perspective, if one writes down the classical intersection multiplicity of two cycles, that can be done by first multiplying both cycles of complementary codim [so for this we need a ring structure, but let's just assume somebody can give such a structure even over Z, just to find out where we would actually be going], then the product lies in CH^n(X), n being the dimension of our scheme. Now to turn that into the classical intersection multiplicity one could pushforward this cycle along the structural map to the base field,

$X$ --> $Spec$ $(k)$

over a field(!) and $CH{\_0}(Spec k) = \mathbb{Z}$ and we get our intersection number. Voilà.
But if we are proper over Spec Z, we could at best pushforward

$X$ --> $Spec$ Z

but $CH{\_0}(Spec(\mathbb{Z}) = 0$, so nothing very interesting seems to result here.
[this argument however only makes sense if the dimension shifts in this Spec Z setup would be carried over analogously, which maybe is also stupid here for the reason that Spec Z is one-dimensional and Spec k zero-dimensional. I am just saying all this, because best and supercool would of course be somebody with a Spec *F*$_1$ having

CH_0(Spec *F*$_1$)= ? (....something, probably rather R than Z)

and that could then be our Spec Z intersection number by giving *F*$_1$ the role of a "virtual base field" and I guess some people say this should link to the Arakelovian one.... but well, that's very speculative]

So I think some people's expectation goes in the direction that the "interesting" way of doing intersection theory over Spec Z needs such a final *F*$_1$-twist.

Note maybe that the classical analgoue would be

P1 <-> Spec Z + (infinite place)

but CH_0(P1) = Z, whereas CH_0(Z) = 0, so we kind of miss something if we just use classical Chow groups over Spec Z. For other questions, classical methods work well even for Z without needing *F*$_1$ or so, for example the étale fundamental groups of both P1 and Spec Z are trivial. But for Chow theory some additional tricks seem to be required.

At least that is my impression. Of course this arithmetic aspect of intersection theory over Spec Z is a kind of different story and it also makes perfect sense to talk about classical Chow groups over Spec Z, so there is certainly nothing wrong in having CH_0(Z) = 0, just maybe for some sorts of questions of arithmetical content, this type of Chow theory may not be the right approach.

The last chapter of Fulton's book defines Chow groups on schemes $X$ of finite type over a fixed smooth scheme $S$. It not automatic to have an intersection product for $X$ smooth, but there is one as long as $S$ is $1$-dimensional. So it should cover the case of $\mathbb{Z}$.

thank you, and what does "smooth scheme S" mean? (smooth is always a relative notion), bc if it means S is smooth over a field k, then it coincides with the above definition. I think the first chapter of 'Lectures on Arakelov geometry' is also a good reference, here they treat schemes over Dedekind rings (for example)
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IvanMar 9 '10 at 18:41

Why always a relative notion? You can just ask for the local rings to be regular.
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Andrea FerrettiMar 9 '10 at 19:07

while smooth over (say) perfect fields implies regular, regularity does not imply smoothness (eg over inseparable extensions). i'm referencing the definition of smoothness (of a morphism) (EGA IV) for arbitrary schemes (not just varieties). Here is a typical example where regular does not imply smooth: let k be an inperfect field, take a finite inseparable extension L/k, then $\mathbb{P}^1 _L$ is regular (variety over k) but not smooth over k.
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IvanMar 9 '10 at 19:26

Ok, I understand what you mean. I do not have Fulton's book right now, but I think he actually means regular.
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Andrea FerrettiMar 9 '10 at 19:43

You mean <i>horizontal</i> divisors ? Following Fulton, one can define a $0$-cycle on ${\rm Spec}(\mathbb Z)$ as intersection. But one can not define its degree as an integer depending only on the class in $CH_0({\mathbb Z})$.
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Qing LiuMar 9 '10 at 21:09